chapt 07 par ti solve

8/2/2019 Chapt 07 Par Ti Solve

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Assembly Language for IntelAssembly Language for Intel--BasedBasedComputers, 4Computers, 4thth EditionEdition

Chapter 7: Integer Arithmetic

(c) Pearson Education, 2002. All rights reserved.

Kip R. Irvine


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Irvine, Kip R. Assembly Language for Intel-Based Computers, 2003. 2

Chapter OverviewChapter Overview

Shift and Rotate Instructions

Shift and Rotate Applications Multiplication and Division Instructions

Extended Addition and Subtraction

ASCII and Packed Decimal Arithmetic


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Shift and Rotate InstructionsShift and Rotate Instructions

Logical vs. Arithmetic Shifts

SHL Instruction

SHR Instruction

SAL and SAR Instructions

ROL Instruction

ROR Instruction

RCL and RCR Instructions

SHLD/SHRD Instructions


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LogicalLogical vsvs Arithmetic ShiftsArithmetic Shifts

A logical shift fills the newly created bit position withzero:

An arithmetic shift fills the newly created bit positionwith a copy of the numbers sign bit:

CF

0

CF


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SHL InstructionSHL Instruction

The SHL (shift left) instruction performs a logical leftshift on the destination operand, filling the lowest bitwith 0.

CF

0

Operand types:SHL reg,imm8

SHL mem,imm8

SHL reg,CLSHL mem,CL


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Fast MultiplicationFast Multiplication

mov dl,5shl dl,1

Shifting left 1 bit multiplies a number by 2

0 0 0 0 1 0 1 0

0 0 0 0 0 1 0 1 = 5

= 10

Before:

After:

mov dl,5shl dl,2 ; DL = 20

Shifting left nbits multiplies the operand by 2n

For example, 5 * 22 = 20


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SHR InstructionSHR Instruction

The SHR (shift right) instruction performs a logicalright shift on the destination operand. The highest bitposition is filled with a zero.

CF

0

mov dl,80

shr dl,1 ; DL = 40shr dl,2 ; DL = 10

Shifting right nbits divides the operand by 2n


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SAL and SAR InstructionsSAL and SAR Instructions

SAL (shift arithmetic left) is identical to SHL.

SAR (shift arithmetic right) performs a right arithmetic

shift on the destination operand.

CF

An arithmetic shift preserves the number's sign.

mov dl,-80sar dl,1 ; DL = -40

sar dl,2 ; DL = -10


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Your turn . . .Your turn . . .

mov al,6Bhshr al,1 a.

shl al,3 b.

mov al,8Ch

sar al,1 c.

sar al,3 d.

Indicate the hexadecimal value of AL after each shift:

35h

A8h

C6h

F8h


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ROL InstructionROL Instruction

ROL (rotate) shifts each bit to the left

The highest bit is copied into both the Carry flagand into the lowest bit

No bits are lost

CF

mov al,11110000b

rol al,1 ; AL = 11100001b

mov dl,3Fh

rol dl,4 ; DL = F3h


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ROR InstructionROR Instruction

ROR (rotate right) shifts each bit to the right

The lowest bit is copied into both the Carry flag and

into the highest bit No bits are lost

CF

mov al,11110000b

ror al,1 ; AL = 01111000b

mov dl,3Fh

ror dl,4 ; DL = F3h


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mov al,6Bhror al,1 a.

rol al,3 b.

Indicate the hexadecimal value of AL after each rotation:

B5h

ADh


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RCL InstructionRCL Instruction

RCL (rotate carry left) shifts each bit to the left

Copies the Carry flag to the least significant bit

Copies the most significant bit to the Carry flag

CF

clc ; CF = 0

mov bl,88h ; CF,BL = 0 10001000brcl bl,1 ; CF,BL = 1 00010000b

rcl bl,1 ; CF,BL = 0 00100001b


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RCR InstructionRCR Instruction

RCR (rotate carry right) shifts each bit to the right

Copies the Carry flag to the most significant bit

Copies the least significant bit to the Carry flag

stc ; CF = 1

mov ah,10h ; CF,AH = 00010000 1

rcr ah,1 ; CF,AH = 10001000 0

CF


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stcmov al,6Bh

rcr al,1 a.

rcl al,3 b.

Indicate the hexadecimal value of AL after each rotation:

B5h

AEh


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SHLD InstructionSHLD Instruction

Shifts a destination operand a given number of bits tothe left

The bit positions opened up by the shift are filled bythe most significant bits of the source operand

The source operand is not affected

Syntax:

SHLD destination, source, count


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SHLD ExampleSHLD Example

.data

wval WORD 9BA6h

.codemov ax,0AC36h

shld wval,ax,4

9BA6 AC36

BA6A AC36

wval AX

Shift wval 4 bits to the left and replace its lowest 4 bits withthe high 4 bits of AX:

Before:

After:


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SHRD InstructionSHRD Instruction

Shifts a destination operand a given number of bits tothe right

The bit positions opened up by the shift are filled bythe least significant bits of the source operand

The source operand is not affected

Syntax:

SHRD destination, source, count


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SHRD ExampleSHRD Example

mov ax,234Bh

mov dx,7654h

shrd ax,dx,4

Shift AX 4 bits to the right and replace its highest 4 bits withthe low 4 bits of DX:

Before:

After:

7654 234B

7654 4234

DX AX


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mov ax,7C36h

mov dx,9FA6h

shld dx,ax,4 ; DX =

shrd dx,ax,8 ; DX =

Indicate the hexadecimal values of each destination operand:

FA67h

36FAh


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Shift and Rotate ApplicationsShift and Rotate Applications

Shifting Multiple Doublewords

Binary Multiplication Displaying Binary Bits

Isolating a Bit String


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Shifting MultipleShifting Multiple DoublewordsDoublewords

Programs sometimes need to shift all bits within anarray, as one might when moving a bitmappedgraphic image from one screen location to another.

The following shifts an array of 3 doublewords 1 bit tothe right (view complete source code):

.data

ArraySize = 3

array DWORD ArraySize DUP(99999999h) ; 1001 1001...

.code

mov esi,0

shr array[esi + 8],1 ; high dword

rcr array[esi + 4],1 ; middle dword, include Carry

rcr array[esi],1 ; low dword, include Carry


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Binary MultiplicationBinary Multiplication

We already know that SHL performs unsignedmultiplication efficiently when the multiplier is a powerof 2.

You can factor any binary number into powers of 2.

For example, to multiply EAX * 36, factor 36 into 32 + 4and use the distributive property of multiplication to

carry out the operation:

EAX * 36

= EAX * (32 + 4)

= (EAX * 32)+(EAX * 4)

mov eax,123

mov ebx,eax

shl eax,5 ; mult by 25

shl ebx,2 ; mult by 22add eax,ebx


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mov ax,2 ; test value

mov dx,ax

shl dx,4 ; AX * 16

push dx ; save for latermov dx,ax

shl dx,3 ; AX * 8

shl ax,1 ; AX * 2

add ax,dx ; AX * 10

pop dx ; recall AX * 16add ax,dx ; AX * 26

Multiply AX by 26, using shifting and addition instructions.Hint:26 = 16 + 8 + 2.


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Displaying Binary BitsDisplaying Binary Bits

Algorithm:Shift MSB into the Carry flag; If CF = 1,append a "1" character to a string; otherwise, append a"0" character. Repeat in a loop, 32 times.

mov ecx,32

mov esi,offset buffer

L1: shl eax,1

mov BYTE PTR [esi],'0'jnc L2

mov BYTE PTR [esi],'1'

L2: inc esi

loop L1


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Isolating a Bit StringIsolating a Bit String

The MS-DOS file date field packs the year, month,and day into 16 bits:

DH DL

Year Month Day

9-15 5-8 0-4

Field:

Bit numbers:

01 000 10 1 10 1 010 10

mov ax,dx ; make a copy of DX

shr ax,5 ; shift right 5 bitsand al,00001111b ; clear bits 4-7

mov month,al ; save in month variable

Isolate the Month field:

chapt 07 par ti solve

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